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I've been reviewing various problems dealing with interesting homeomorphisms, and I came across this one.

Is the product of the space of irrationals and the space of rationals homeomorphic to the space of irrationals?

I haven't been able to make any progress on this one. Can anyone help? Thank you!

Maria
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  • What topologies are given for the irrationals and the rationals? The subspace topology from their inclusions in the real line? – Shaun Ault Sep 02 '12 at 21:38
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    @Shaun If someone doesn't say what topology a subset of $\Bbb R$ has, you should assume it's the usual one. – MJD Sep 02 '12 at 21:49
  • @ShaunAult : Those are linearly ordered sets and there's an order topology. I'm pretty sure that's the same as the subspace topology that you mention, but it seems simpler to put it that way. – Michael Hardy Sep 02 '12 at 22:13

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Let $\Bbb P$ be the space of irrationals; it is topologically complete, meaning that it has a compatible complete metric. Suppose that $\Bbb P\times\Bbb Q$ were homeomorphic to $\Bbb P$; then it would be topologically complete, so each of its closed subspaces would also be topologically complete. But $\Bbb P\times\Bbb Q$ certainly has closed subspaces homeomorphic to $\Bbb Q$, which is not topologically complete, so $\Bbb P\times\Bbb Q\not\cong\Bbb P$.

Added: In this paper Jan van Mill proved that $\Bbb P\times\Bbb Q$ is the unique space (up to homeomorphism) that can be written as an increasing union $\bigcup_{n\in\Bbb N}F_n$ of closed sets such that for each $n\in\Bbb N$, $F_n$ is a copy of $\Bbb P$ that is nowhere dense in $F_{n+1}$. (To write $\Bbb P\times\Bbb Q$ this way, enumerate $\Bbb Q=\{q_n:n\in\Bbb N\}$, and let $F_n=\Bbb P\times\{q_k:k\le n\}$.) It’s a nice little exercise in the Baire category theorem to show that $\Bbb P$ cannot be written in this way.

Brian M. Scott
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    A slightly different way of putting it: If $\mathbb{P}$ were homeomorphic to $\mathbb{P} \times \mathbb{Q} = \bigcup_{q \in \mathbb{Q}} \mathbb{P} \times {q}$ it would be a countable union of nowhere dense sets, contradicting Baire. – t.b. Sep 02 '12 at 21:52
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    @t.b.'s remark provides a proof that doesn't require showing that $\mathbb P$ is topologically complete: all you have to note is that if $\mathbb P$ was a countable union of nowhere dense closed (in $\mathbb P$) sets $S_j$, ${\mathbb R}$ would be as well. Just take the closures (in $\mathbb R$) of $S_j$ and the singletons ${q}$ for $q \in \mathbb Q$. – Robert Israel Sep 02 '12 at 22:01
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    Is it clear that $\Bbb P$ is topologically complete? I didn't know that, and I can't think of a proof offhand. – MJD Sep 02 '12 at 22:01
  • @t.b. That should go on an answer rather than a comment! – Asaf Karagila Sep 02 '12 at 22:06
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    @MJD: It’s well enough known that it’s part of my mental furniture. It follows from the fact that a subset of a complete metric space $X$ is topologically complete iff it’s a $G_\delta$ in $X$, which is standard fare. – Brian M. Scott Sep 02 '12 at 22:09
  • @MJD: Two ways: 1) use continued fractions to show that $\mathbb{P}$ is homeomorphic to $\mathbb{Z} \times \mathbb{N}^{\mathbb{N}}$. 2) It is a general fact that a subspace of a complete metric space is topologically complete if and only if it is a countable intersection of open sets. I outline an argument of topological completeness of $G_\delta$'s in a comment here. Necessity is harder and uses Lavrentiev's theorem which I discuss here – t.b. Sep 02 '12 at 22:10
  • @t.b.: You don’t even need continued fractions: you can do it by a fairly straightforward inductive construction that’s well described in these notes by Arnie Miller (Theorem 1.1). – Brian M. Scott Sep 02 '12 at 22:19
  • Thanks, but that argument doesn't fit easily into a comment box :) I knew the argument (I always wanted to know what the reals were, anyway!). The published version of these lecture notes (and volumes 1-12 of the Springer lecture notes in logic are freely available on the Project Euclid site). – t.b. Sep 02 '12 at 22:29
  • @t.b.: I thought you might. I remember sitting down and working out something very similar when I first learned that the irrationals are homeomorphic to $\omega^\omega$. – Brian M. Scott Sep 02 '12 at 22:41